Now, not all tricky limits are in the form 0 over 0.

For example, the limit as x goes to 0 of 1 over sine squared x minus 1 over x

squared. Evaluating looks like infinity minus

infinity. That's not going to work.

So, let us put this over the common denominator of x squared times sine

squared of x. The resulting numerator is x squared

minus sine squared of x. Now we can apply L'Hopital's rule, or in

this case, Taylor expansion. If we Taylor expand, well, the only thing

we need to worry about is sin squared of x, which is the Taylor series for sin of

x, quantity squared. What we really want to do is compute the

leading order term. So we're going to have a little bit of

algebra to do when we square x minus x cubed over 3 factorial plus higher order

terms, what do we get? The first term is x squared, the second

term is 2 times x times x cubed over 3 factorial.

Now, what do we note here? We note that the x squareds cancel.

Those second ordered terms, they go away and we're left with the next term.

The fourth order term as the leading order coefficient.

Likewise, in the denominator, we see that the leading order term is simply x to the

4th, thus cancelling the x to the 4ths gives us leading order coefficients of 1

3rd in the numerator and one in the denominator.

Our answer is 1 3rd. Likewise, if we consider x times log of x

and send x to 0. We're going to have to be a little

careful. This is a one-sided limit, sending x to 0

from the right. Then, what would this be?

While x is going to 0, log of x is going to minus infinity.

0 times minus infinity, that's not going to work.

Once again, let us put this over a denominator.

In this case, we'll move the x to the denominator and call it 1 over x.

So that we have a ratio of two functions. In this case, log of x over x to the

minus 1. Now, we're going to apply L'Hopital's

Rule. I'm a little nervous about Taylor

expansions here, because both log of x and 1 over x, don't seem to have good

Taylor expansions at x equals 0. So, applying L'Hopital's Rule gives us

what? The derivative of log of x is 1 over x.

The derivative of 1 over x is minus 1 over x squared.

These simplify to minus x and taking the limit as x goes to 0 from the right, we

see that the answer is 0. Now, L'Hopital's rule can also help in

evaluating limits at infinity. Consider the limit as x goes to infinity.

Of log of x over square root of x. It's maybe not apparent what that limit

is going to be. Log of x and square root of x, both

become infinite as x goes to infinity. But at what rate?

Is there one that is bigger than the other?

Well, let's apply L'Hopital's Rule. This gives us the limit as x goes to

infinity of what? The derivative log of x is 1 over x.

The derivative of x to the one half is one half x to the minus one half.

This simplifies to twice x to the minus 1 half as x goes to infinity, this clearly

goes to 0. Well, that's the answer.

But what does that really mean? A log of x is going to infinity, square

root of x is going to infinity, their ratio is going to 0.

What that really means is that the square root of x dominates the log of x.

It grows much faster than the logarithm does.